Kites & kiting

[Turtle]

Thank you thank you! Now for using C1 to find X =XK I think.

 

 

 

 

 

So that last result, i.e. 11995.047124752534496020791375179, I expected to be the final answer, that is it should be the altitude of the kite in feet. Did I do that right then?

 

Hold it! X = XK is the distance along the ground, not the altitude which is Y. That's why I originally thought I did something wrong. Here's the NASA page I used. >> Control Line Equations

 

I left out of my earlier quote my uncertainty with C2. Here it is too.

 

 

EDIT: It's possible the minus sign before (D/P) is a typo. I've seen a couple already. :unsure:

 

C2 = -(D/p) * cosh (C1)

= -(11.52/0.0011733) * cosh(0.5016146144159724309580791536331)

= -9818.4607517259013040143185885963 * 1.1284688034136246239705274547244

= -11079.826655863765165039185441426

 

 

or is it...

 

C2 = -(D/p) * cosh (C1)

= (-11.52/-0.0011733) * cosh(0.5016146144159724309580791536331)

= 9818.4607517259013040143185885963 * 1.1284688034136246239705274547244

= 11079.826655863765165039185441426

 

 

Thennnnn...also if I got C1 correct I use it to calculate XK.

 

XK = (D / p) * [sinh^-1((L - W) / D) - C1]

= (11.52/ 0.0011733) * [sinh^-1((25.624 - 2) / 11.52) - 0.5016146144159724309580791536331]

= 9818.4607517259013040143185885963 * sinh^-1(2.0506944444444444444444444444444 - 0.5016146144159724309580791536331)

= 9818.4607517259013040143185885963 * sinh^-1(1.5490798300284720134863652908113)

= 9818.4607517259013040143185885963 *1.2216830548152906878629509132376

= 11995.047124752534496020791375179

 

:coffee_n_pc:

 

 

 

 

Edited September 26, 2013 by Turtle

[phillip1882]

umm no, p = pressure = 14.674 roughly, so D/p = 0.7839 roughly.

then you have your parenthesis wrong. it should be 0.7839*(sinh^-1(2.05069) -.501)

[Turtle]

umm no, p = pressure = 14.674 roughly, so D/p = 0.7839 roughly.

then you have your parenthesis wrong. it should be 0.7839*(sinh^-1(2.05069) -.501)

 

The diagram says p is the line weight per foot. ??. I just edited my last post so you better check that too. Again; thank you.

Edited September 26, 2013 by Turtle

[phillip1882]

if p = line weight per length then g != 0.0011733, that's p.

therefore C1 is wrong.

edit: with the following i get:

import math
p= 0.001173333
W = 2 #oz for plastic kite and 1/4" birch sticks
D = 11.52
L = 25.624
s = 1760.0
g = p*s
C1 = math.asinh((L-g-W)/D)
C2 = - D/p * math.cosh(C1)
X = 0.756145664696
y = C2 + D/p* math.cosh(p/D *X +C1)
print y

1.41513942445

Edited September 26, 2013 by phillip1882

[Turtle]

import math
g= 0.001173333
W = 1.773 #oz for plastic kite and 1/4" birch sticks
D = 11.52
L = 25.624
p = 14.694

y = (- D/p * math.cosh(math.asinh((L-g-W)/D))
+ D/p * math.cosh(p/D * (math.asinh((L-W)/D) 
-math.asinh((L-g-W)/D))) + math.asinh((L-g-W)/D))
print y

outputs 0.456106520742

and i can confirm your result for C1

 

Uhmmm...g should be 17.6 oz, the total weight of the line. You seem to have changed it to the weight per foot, which is p.

 

Confusing; ain't it?

[Turtle]

if p = line weight per length then g != 0.0011733, that's p.

therefore C1 is wrong.

 

OK. The page says C1 = sinh^-1[(L - g - W) / D ]

 

L = lift and I got 25.624oz from the modeler

 

W= weight of kite and I got 1.773oz from the modeler

 

g= total weight of line and I got that from Double-D

 

D = drag and I got 11.52oz from the modeler

 

C1 = ???

[phillip1882]

okay fixes:

import math
p= 0.001173333
W = 2 #oz for plastic kite and 1/4" birch sticks
D = 11.52
L = 25.624
s = 1760.0
g = p*s
C1 = math.asinh((L-g-W)/D)
C2 = - D/p * math.cosh(C1)
X = D/p*(math.asinh((L-W)/D) -C1)
y = C2 + D/p* math.cosh(p/D *X +C1)

print y

gives 1567.64820561

[Turtle]

okay fixes:

import math
p= 0.001173333
W = 2 #oz for plastic kite and 1/4" birch sticks
D = 11.52
L = 25.624
s = 1760.0
g = p*s
C1 = math.asinh((L-g-W)/D)
C2 = - D/p * math.cosh(C1)
X = D/p*(math.asinh((L-W)/D) -C1)
y = C2 + D/p* math.cosh(p/D *X +C1)

print y

gives 1567.64820561

 

Nope. s = total line length which Double-D gives as 15000 feet. I expect a correct Y, i.e. the kite's altitude should be around 10000 ft. Note Double-D actually weighed the kite at about 2 oz, but the modeler gave it as 1.7 oz. using its own estimate of the materials used, that is plastic sails and wooden sticks.

 

Have a look at the diagram in case I'm messing up something in my translation. There is a lot of explanatory text at the page with the diagram. The link is just above the diagram I just posted.

[phillip1882]

okay then, allowing s to be 15000 gives 11320.89 feet

[Turtle]

okay then, allowing s to be 15000 gives 11320.89 feet

 

That's in the range I expected. 2.1 miles up then. Pretty impressive for a guy in his backyard with a dime-store kite and fishing line; but then we already knew that. The scientists down on Christmas island I earlier referenced were bragging about the same achievement with expensive parafoils and Kevlar line. :lol:

 

...

Article is...well, as articles do. [bolding in quote is mine.]

Ben Balsley, CIRES Scientist in the Field Gathering atmospheric dynamics data using KITES

[Reprinted from Summit Magazine, Spring 1994]...

The researchers launched the tandem kite system -- which was spooled by a hand-powered winch -- from an abandoned runway near the southern edge of the shoe-shaped island. The 2.2-mile height attained by the upper kite proved to be a new world record for parafoils. "Once it was up it just stayed there," he marvels. "It became obvious there is a natural niche for research kites." ...

 

Anyway, I still don't know where I was messing up and I'd like to be able to do this calculation myself for my own future flights. Would it be too great an imposition for you to show the work step by step? Thanks again Phillip. You da man. :bouquet:

 

C1 =

 

C2 =

 

XK =

 

Y=

Edited September 26, 2013 by Turtle

[phillip1882]

i have each calculation in the code.

but I'll re post in more math like terms.

p= 0.001173333

W = 2

D = 11.52

L = 25.624

s = 15000.0

g = p*s

 

C1 = sinh^-1((L-g-W)/D)

C2 = - D/p *cosh(C1)

XK = D/p*(sinh^-1((L-W)/D) -C1)

y = C2 + D/p* cosh(p/D *XK +C1)

Edited September 26, 2013 by phillip1882

[Turtle]

i have each calculation in the code.

but I'll re post in more math like terms.

p= 0.001173333

W = 2

D = 11.52

L = 25.624

s = 15000.0

g = p*s

 

C1 = sinh^-1((L-g-W)/D)

C2 = - D/p *cosh(C1)

XK = D/p*(sinh^-1((L-W)/D) -C1)

y = C2 + D/p* cosh(p/D *XK +C1)

 

Well, I had all that already. What I meant by showing each step-by-step is to give each equation calculation in the form that I gave like this:

 

C1 = sinh^-1[(L - g - W) / D ]

= sinh^-1[(25.624 - 17.6 - 2) / 11.52]

= sinh^-1[6.024/11.52]

= sinh^-1[0.5229166]

= 0.5016146144159724309580791536331

 

At one point you said my C1 was wrong and at another point you said you confirmed my C1 but you gave 'outputs 0.456106520742' which is different than my 0.5016... I'm using pencil & paper and a simple calculator. Well, you know me and my methods.

 

If it's a bother, no worry. I have plenty of time before I'm flying again to try & muddle my way through this one specific case. Thanks again. :)

Edited September 26, 2013 by Turtle

[LaurieAG]

g= total weight of line and I got that from Double-D

 

There is a different definition of g on the Control Line Equations page below the image.

 

http://www.grc.nasa.gov/WWW/K-12/airplane/kitesag.html

 

The control line has a certain length s and the weight of the control line is evenly distributed along the length at p weight per length of line. The total weight of the control line is designated g and is given by:

 

g = s * p

[Turtle]

g= total weight of line and I got that from Double-D

There is a different definition of g on the Control Line Equations page below the image.

http://www.grc.nasa....ne/kitesag.html

 

g=s*p

 

Thanks Laurie. :bouquet: Confusing enough without me isn't it? :lol: I later gave the correct assignment even though I missed that mistake.

Nope. s = total line length which Double-D gives as 15000 feet. I expect a correct Y, i.e. the kite's altitude should be around 10000 ft.

 

Doble-Dee originally said his 5000 yds of line weighed 1.1 pounds. Looks like Phillip has it correct in spite of me in his last summary. Do you agree?

 

Did I see somewhere a regulation that you Aussies can't fly above 250 ft? Or maybe that's the Brits? Since my Fall is fallen, your Spring must have sprung. Doing any flying yourself yet? I think I saw some new power generation schemes in my last couple months of reading; I'll review what we discussed here and see if there's something of interest to add.

 

Thanks again for all the help fellas! You're hale & well met. :wave: :circle:

[DFINITLYDISTRUBD]

WOW! Y'all's been busy! Mighty impressive!

I note that the line length does not take into account 1" per foot (give or take a millimeter or so) of stretch I determined for the line under load

making 15,000 feet nearly 1250 feet off (doesn't really matter, just havin a giggle).

Am currently gearing up for a 7,000-10,000 yard attempt, having found the perfect publicly accessible field immediately adjacent to the lake with uninterrupted expanses to the east, which oddly enough is the most common wind direction here and generally the direction of the strongest breezes....first I need to sort a reliable powered reel....No friggen way I'm reeling in 4 - 5.7 miles of line manually

 

Ok enough of my interuption, back to the show....

[Turtle]

WOW! Y'all's been busy! Mighty impressive!

I note that the line length does not take into account 1" per foot (give or take a millimeter or so) of stretch I determined for the line under load

making 15,000 feet nearly 1250 feet off (doesn't really matter, just havin a giggle).

 

Thnx 4 da props. :) Giggle away.Line stretch is no small matter though. Using this theoretical method we obtain at best an estimate. Besides line stretch and differing line drag through different wind layers, at such high altitudes the kite looses lift in the thinner air and this is no small matter either. The triangulation method Laurie & I mentioned is a bit better, but no less troublesome getting a decent read at such height. The fellas down at Christmas Island had an onboard altimeter; the best possible method for establishing a kite's altitude.

 

Am currently gearing up for a 7,000-10,000 yard attempt, having found the perfect publicly accessible field immediately adjacent to the lake with uninterrupted expanses to the east, which oddly enough is the most common wind direction here and generally the direction of the strongest breezes....first I need to sort a reliable powered reel....No friggen way I'm reeling in 4 - 5.7 miles of line manually

 

Ok enough of my interuption, back to the show....

 

Yeah; like it's an interruption to post on-topic. Lima Charlie Lima Now the reel work begins. ;)

[DFINITLYDISTRUBD]

Well fine and dandy then!

 

Been thinkin about them carriages....If only I could afford a couple spools of 40lb....two spool rig with a quick release at the end of spool one and a quick connect at the beginning of spool two....spool one pays out from the spool to a pulley on the kite to a quick connect on the spool holder...at it's end the carriage is connected to the line AND the quick connect and is carried away...upon the carriage reaching the kite the tail of line one is connected to the quick connect on line two...line tow pays out....kite goes very high and very far....camera shoots really groovy video...time for kite to come down...start motor on spool...suck up all of line two...reconnect line one to line one's spool....suck up all of line one...finish beer...pack up...go home...watch video...much celebration.